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Chapter 8: Problem 24

A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. \(8-37\) ). Find the maximum distance the spring is compressed.

Short Answer

Expert verified
The spring compresses approximately 0.089 meters.

Step by step solution

01

Convert Height to Meters

First, convert the height from centimeters to meters since standard units are necessary for the calculations. Height: \( h = 40 \text{ cm} = 0.40 \text{ m} \).
02

Identify Energy Transformation

As the block falls, potential energy due to height is converted to potential energy stored in the spring. This transformation is governed by conservation of energy principles.
03

Write Potential Energy Equations

Initially, the gravitational potential energy at the beginning is \( U_g = mgh \) and when the spring is completely compressed, all energy is stored as spring potential energy \( U_s = \frac{1}{2}kx^2 \), where \( x \) is the spring compression.
04

Set up Conservation of Energy Equation

Set up the equation: the initial gravitational potential energy equals the spring potential energy at maximum compression. This gives: \[ mgh = \frac{1}{2}kx^2 \].
05

Insert Values into the Equation

Substitute the known values into the energy equation: \( (2.0 \times 9.8 \times 0.40) = \frac{1}{2}(1960)x^2 \).
06

Solve for Maximum Spring Compression

Simplify and solve the equation: \( 7.84 = 980x^2 \). Divide both sides by 980 to isolate \( x^2 \): \( x^2 = \frac{7.84}{980} \).
07

Simplify and Compute Final Answer

Compute \( x^2 = 0.008 \), then take the square root to find \( x \): \( x = \sqrt{0.008} \approx 0.089 \). Thus, the spring is compressed by approximately \( 0.089 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy is a form of energy that is stored within an object due to its position, arrangement, or state. It has the potential to be converted into other forms of energy, such as kinetic energy. In this exercise, when the block is at a height of 40 cm above the ground, it possesses gravitational potential energy due to its elevated position relative to the ground.
  • Gravitational potential energy is calculated using the formula: \( U_g = mgh \), where \( m \) represents the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)), and \( h \) is the height above the reference point.
  • As the block is released and begins to fall, this potential energy decreases as it is converted into kinetic energy and eventually into the spring's potential energy when the block makes contact with the spring.
Spring Compression
Spring compression occurs when a force is applied to a spring, causing it to shorten. In the context of our problem, as the block falls onto the spring, it depresses鈥攐r compresses鈥攖he spring.
  • The amount of compression in the spring is directly related to the energy transferred to it by the falling block.
  • As the spring compresses, it stores energy, which in this case was initially present as gravitational potential energy in the block.
Using the formula for potential energy stored in a spring, \( U_s = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the spring's displacement (compression), we can understand how the spring captures this transferred energy.
As per the conservation of energy, the initial gravitational potential energy must equal the spring's potential energy at maximum compression.
Gravitational Potential Energy
Gravitational potential energy is the energy held by an object because of its position in a gravitational field. This energy becomes particularly important when dealing with objects that change altitudes, such as the block in our exercise.
  • The formula \( U_g = mgh \) quantifies this energy, where \( m \) stands for mass, \( g \) is gravitational acceleration, and \( h \) is height.
  • As the block is dropped, its gravitational potential energy decreases as it falls downward, transforming into kinetic energy and then into energy stored in the compressed spring.
This transformation underscores the principle of conservation of energy: energy in an isolated system remains constant but can change forms. At the peak compression of the spring, all gravitational potential energy has converted into spring potential energy.
Spring Constant
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It indicates how much force is needed to compress (or extend) the spring by a certain distance and is measured in Newtons per meter (\( \text{N/m} \)).In our problem, the spring constant is given as \( k = 1960 \text{ N/m} \). This high spring constant suggests that the spring is very stiff, requiring considerable force to compress it.
  • The relationship between spring compression and force is described by Hooke's Law: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring.
  • A higher spring constant means the spring is harder to compress or extend.
The spring's constant plays a crucial role in determining how much the spring will compress when the block hits it, helping us understand the transfer of energy from the falling block to the spring.

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Most popular questions from this chapter

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-42). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm}\), is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

Ice flake is released from the edge of a hemispherical bowl whose radius \(r\) is \(22.0\) \(\mathrm{cm}\). The flake-bowl contact is frictionless. (a) How much work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl? (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (d) increase, decrease, or remain the same?

A \(68 \mathrm{~kg}\) sky diver falls at a constant terminal speed of \(59 \mathrm{~m} / \mathrm{s}\). (a) At what rate is the gravitational potential energy of the Earth-sky diver system being reduced? (b) At what rate is the system's mechanical energy being reduced?

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m}\). The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0\) \(\mathrm{cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?
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